# The closest sum of three numbers of Leetcode-16

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Similar topics: The sum of three numbers of Leetcode-15

Sort + double pointer

Sort the array nums, lock a number nums[i], and then use the double pointer to consider the sum of the remaining two numbers
Dual pointer initialization: LF = i + 1 (no need to start from scratch, if there is this combination, it will be found when i is smaller)
For each value of ans, the absolute difference between res and ans and target is smaller

```class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(),nums.end());
int res=INT_MAX/2;
int lf,rt;
for(int i=0;i<nums.size();i++)
{
lf=i+1;
rt=nums.size()-1;
while(lf<rt)
{
int ans=nums[i]+nums[lf]+nums[rt];
if(ans==target)
{
return target;
}
else if(ans<target)
{
res=fabs(res-target)>fabs(ans-target)?ans:res;
lf++;
}
else
{
res=fabs(res-target)>fabs(ans-target)?ans:res;
rt--;
}
}
}
return res;
}
};
```

Consider weight removal

For the de duplication of nums[i]
When i is small, the remaining two numbers are considered more widely.
For example, for the [1,1,1,2,3], target=7 group, for the first 1, the most appropriate [2,3] will be found from [1,1,2,3]. The second 1 and the third 1 are only the search scope is decreasing, so we can consider de duplication.

De duplication of nums[lf] and nums[rt]
Similarly, suppose the input changes to [1,1,1,3,3,3], target=6
For the first 1, the first group found in [1,1,3,3,3] is nums[lf]=1, nums[rt]=3
After that, the search range is tightened inward, but if the nums[lf] or nums[rt] value is the same as the last time, then it is unnecessary to add and sum again for ans, and only the changed value can bring more close results

```class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(),nums.end());
int res=INT_MAX/2;
int lf,rt;
for(int i=0;i<nums.size();i++)
{
if(i>0&&nums[i-1]==nums[i])
continue;
lf=i+1;
rt=nums.size()-1;
while(lf<rt)
{
int ans=nums[i]+nums[lf]+nums[rt];
if(ans==target)
return target;
res=fabs(res-target)>fabs(ans-target)?ans:res;
if(ans<target)
{
lf++;
while(lf<rt&&nums[lf]==nums[lf-1])
lf++;
}
else
{
rt--;
while(lf<rt&&nums[rt]==nums[rt+1])
rt--;
}
}
}
return res;
}
};
```

Posted on Thu, 04 Jun 2020 13:27:00 -0400 by Lashiec